DE_MF

7.Groups may wrap around the ta The leftmost cell in a row may b and the top cell in a column may

able.be grouped with the rightmost celly be grouped with the bottom cell. 31

8. There should be as few gro as this does not contradict

oups as possible, as longany of the previous rules. 32

Summary:• No zeros allowed.• No diagonals.• Only power of 2 number of c• Groups should be as large a• Every one must be in at lea• Overlapping allowed.• Wrap around allowed.• Fewest number of groups p

cells in each group. as possible.ast one group.possible. 33

Example 1:Consider the following map. The funcNote that values of the input variablThat is the logic values of the variabform and zero denoting false formcolumns respectively.

ction plotted is: Z = f(A,B) = A +AB. les form the rows and columns. bles A and B (with one denoting truem) form the head of the rows and 34

The map displayed is a onebe used to simplify an expressiThere is a two-dimensional mto four variables, and a three-dvariables.

e dimensional type which can ion in two variables.map that can be used for updimensional map for up to six 35

 Referring to the map abov grouped together. Through inspection it can b true and false form within th This eliminates variable B l only has its true form. The minimized answer the

ve, the two adjacent 1's arebe seen that variable B has itshe group. leaving only variable A whicherefore is Z = A. 36

Exam• Given function, F = Σ (1, 2,• Since the biggest number i defined by 3 variables.• Let’s draw K-Map for this that are present in function don’t necessarily need to w them.

mple-2 3, 4, 5, 6) in this function is 6, it can be function by writing 1 in cells and 0 in rest of the cells. You write 0s but it is okay to have 37



38

 We need to apply rules for simp tutorial. So, first we need to look f is none, so we should now look f there is none, so we will look for pa (1,3) – A’C (Since B is the changi it is eliminated) (2,6) – BC’ (Since A is the changin (4, 5) – AB’ (Since C is the changin Thus, F = A’C + BC’ + AB’

plifying K-Map that we read in lastfor an octet i.e. 8 adjacent 1′s. Therefor a quad i.e. 4 adjacent 1′s. Again,airs. There are 3 pairs circled in red.ing variable between these two cells,ng variable, it is eliminated)ng variable, it is eliminated) 39

Exam• Given function, F = Σ (0, 4,• Since, the biggest number variables to define this func• Let’s draw K-Map for this that are present in function

mple-3 6, 8, 10, 15) r is 15, we need to have 4 ction. function by writing 1 in cells and 0 in rest of the cells. 40



41

 Applying rules of simplifying K-Map, are 3 pairs, circled in red. (0, 4) – A’C'D’ (Since B is the chang is eliminated) (4, 6) – A’BD’ (Since C is the changin (8, 10) – AB’D’ (Since C is the chang There is 1 in cell 15, which can not b it can not be simplified further and lef 15 = ABCD Thus, F = A’C'D’ + A’BD’ + AB’D’ +